It is known that given a set of Areas $A_f$ and normals $\vec{n}_f$ if $\sum_f A_f \vec{n}_f=0$ exist a unique convex polyhedron with given face areas and normals. (Minkowski theorem - See Alexandrov book on Convex Polyhedra).

Obviously here I'm identifying all the isometric polyhedra.

In principle with the same set of areas and normals one can build "others" polyhedra if we relax the convexity requirement.

What I want to prove is that in the collection of all the possible polyhedra one can build from a given set of Areas $A_f$ and normals $\vec{n}_f$ the convex one is the one with bigger volume.

One could ask a more general question. Consider a measure on $S^2$, and suppose this is the push-forward of the area measure of a surface in $R^3$ under the Gauss map. Then one can ask whether a convex surface realizing this Gauss map measure will have larger volume than a non-convex surface with the same Gauss map measure.
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Ian AgolJul 5 '13 at 15:27

There is in fact an extension of Minkowski thm to non-convex polyhedra. See G.Y. Panina tinyurl.com/n9cpbvy (MR1970337) and V. Alexandrov arxiv.org/pdf/math/0211286v1.pdf This does not really answer your question, but suggests a possibility that the claim might be false.
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Igor PakJul 7 '13 at 6:18

I believe this was how Minkowski proved his theorem (he used a variational principle to show that the sought-after polyhedron maximizes volume). For the proof, see Aleksandrov's "Convex polyhedra", section 7.2